Finite Math Examples

$20 x + 20 x = 15.2$ , $20 x + 12 x = 14.52$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Add $20 x$ and $20 x$ .

$40 x = 15.2$

$20 x + 12 x = 14.52$

Step 1.2

Add $20 x$ and $12 x$ .

$40 x = 15.2$

$32 x = 14.52$

$40 x = 15.2$

$32 x = 14.52$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 40 & 15.2 \\ 32 & 14.52 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{40}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{40}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{40}{40} & \frac{15.2}{40} \\ 32 & 14.52 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0.38 \\ 32 & 14.52 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 32 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 32 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 0.38 \\ 32 - 32 \cdot 1 & 14.52 - 32 \cdot 0.38 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0.38 \\ 0 & 2.36 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{1}{2.36}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{2.36}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0.38 \\ \frac{0}{2.36} & \frac{2.36}{2.36} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0.38 \\ 0 & 1 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} - 0.38 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} - 0.38 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 0.38 \cdot 0 & 0.38 - 0.38 \cdot 1 \\ 0 & 1 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 4

Use the result matrix to declare the final solution to the system of equations.

$x = 0$

$0 = 1$

Step 5

The system is inconsistent so there is no solution.

$No solution$